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प्रश्न
Maximize Z = 3x + 5y
Subject to
\[x + 2y \leq 20\]
\[x + y \leq 15\]
\[ y \leq 5\]
\[ x, y \geq 0\]
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उत्तर
We need to maximize Z = 3x + 5y
First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 20, x + y = 15, y = 5 , x = 0 and y = 0.
The line x + 2y = 20 meets the coordinate axis at A(20, 0) and B(0,10). Join these points to obtain the line x + 2y = 20.
Clearly, (0, 0) satisfies the inequation x + 2y ≤ 20. So, the region in xy-plane that contains the origin represents the solution set of the given equation.
The line x + y = 15 meets the coordinate axis at C(15, 0) and D(0,15). Join these points to obtain the line x + y = 15.
Clearly, (0, 0) satisfies the inequation x + y ≤ 15. So, the region in xy-plane that contains the origin represents the solution set of the given equation.
y = 5 is the line passing through (0, 5) and parallel to the X axis.The region below the line y = 5 will satisfy the given inequation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are O(0, 0), \[C\left( 15, 0 \right)\] , \[E\left( 10, 5 \right)\] and \[F\left( 0, 5 \right)\] The values of Z at these corner points are as follows.
| Corner point | Z = 3x + 5y |
| O(0, 0) | 3 × 0 + 5 × 0 = 0 |
|
\[C\left( 15, 0 \right)\]
|
3 × 15 + 5 × 0 = 45
|
|
\[E\left( 10, 5 \right)\]
|
3 × 10 + 5 × 5 = 55 |
|
\[F\left( 0, 5 \right)\]
|
3 × 0 + 5 × 5 = 25 |
We see that the maximum value of the objective function Z is 55 which is at \[E\left( 10, 5 \right)\]
Thus, the optimal value of Z is 55.
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